Again, we follow the proof of Proposition 3.2 in [1]}. We obtain
\(\begin{split}\mathcal {L}^{\nu _h}[\psi ](0)&\ge R^{\alpha -2}\int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)-\psi (0)\int _{|z|>R}\,\mathrm {d}\nu _h(z)\\&\ge R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)+R^\alpha \int _{|z|>R}\,\mathrm {d}\nu _h(z).\end{split}\) 
Hence, since \(R>1\)  and \(\alpha \in (0,2)\) , we have \(R^{\alpha -2}\le 1\)  and
\(\begin{split}&\sup _{0<h<1}\bigg \lbrace R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)+R^\alpha \int _{|z|>R}\,\mathrm {d}\nu _h(z)\bigg \rbrace \\&\le \sup _{0<h<1}\mathcal {L}^{\nu _h}[\psi ](0)\le |\mathcal {L}^\mu [\psi ](0)|+\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|\\&\le \bigg (R^{\alpha -2}\int _{|z|\le 1}|z|^2\,\mathrm {d}\mu (z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)+\int _{|z|>R}|\psi (z)-\psi (0)|\,\mathrm {d}\mu (z)\bigg )\\&\quad +\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|\\&\le \bigg (\int _{|z|\le 1}|z|^2\,\mathrm {d}\mu (z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)+2R^\alpha \int _{|z|>R}\,\mathrm {d}\mu (z)\bigg )\\&\quad +\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|,\end{split}\) 
which is finite due to the pointwise convergence and the assumption (REF ).
